Question #25252

Show that from: rad (R[t]) = (Nil*R)[t] follows Kothe’s Conjecture. (“The sum of two nil left ideals in any ring is nil”.)

Expert's answer

We shall prove Kothe’s Conjecture inthe alternative form: If *R *is a ring such that Nil**R *= 0, thenevery nil left ideal *I **⊆** R *is zero. Itsuffices to show that, for any* f*(*t*) *∈** I*[*t*], 1 + *f*(*t*) has a leftinverse in *R*[*t*], for then *I*[*t*] *⊆** *rad *R*[*t*], and rad *R*[*t*] = 0 by assumption.Consider the ring *R' *= *I **⊕** *Z, obtained by formally adjoining anidentity 1*' *= (0*, *1) to the “rng” *I*. It is easy to seethat Nil**R' *= *I*, so by assumption, rad *R'*[*t*] = *I*[*t*].Given *f*(*t*) *∈** I*[*t*],1*' *+ *f*(*t*) has then a left inverse, say 1*' *+ *g*(*t*),in *R'*[*t*]. From1*' *= (1*' *+ *g*(*t*))(1*' *+ *f*(*t*))*,*we have

*−g*(*t*) = *f*(*t*) + *g*(*t*)*f*(*t*)*∈** **I*[*t*]*,*

since*I*[*t*] is an idealin *R'*[*t*]. Now the equation 1 = (1+*g*(*t*))(1 + *f*(*t*))implies that 1 + *f*(*t*) has a left inverse 1 + *g*(*t*) *∈** R*[*t*], as desired.

since

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